$ E = \left[\begin{array}{rr}-1 & 2 \\ -2 & 3\end{array}\right]$ $ C = \left[\begin{array}{rrr}4 & -2 & 3 \\ 1 & 0 & -2\end{array}\right]$ What is $ E C$ ?
Answer: Because $ E$ has dimensions $(2\times2)$ and $ C$ has dimensions $(2\times3)$ , the answer matrix will have dimensions $(2\times3)$ $ E C = \left[\begin{array}{rr}{-1} & {2} \\ {-2} & {3}\end{array}\right] \left[\begin{array}{rrr}{4} & \color{#DF0030}{-2} & \color{#9D38BD}{3} \\ {1} & \color{#DF0030}{0} & \color{#9D38BD}{-2}\end{array}\right] = \left[\begin{array}{rrr}? & ? & ? \\ ? & ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rrr}{-1}\cdot{4}+{2}\cdot{1} & ? & ? \\ ? & ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rrr}{-1}\cdot{4}+{2}\cdot{1} & ? & ? \\ {-2}\cdot{4}+{3}\cdot{1} & ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rrr}{-1}\cdot{4}+{2}\cdot{1} & {-1}\cdot\color{#DF0030}{-2}+{2}\cdot\color{#DF0030}{0} & ? \\ {-2}\cdot{4}+{3}\cdot{1} & ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rrr}{-1}\cdot{4}+{2}\cdot{1} & {-1}\cdot\color{#DF0030}{-2}+{2}\cdot\color{#DF0030}{0} & {-1}\cdot\color{#9D38BD}{3}+{2}\cdot\color{#9D38BD}{-2} \\ {-2}\cdot{4}+{3}\cdot{1} & {-2}\cdot\color{#DF0030}{-2}+{3}\cdot\color{#DF0030}{0} & {-2}\cdot\color{#9D38BD}{3}+{3}\cdot\color{#9D38BD}{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rrr}-2 & 2 & -7 \\ -5 & 4 & -12\end{array}\right] $